Ncert Solutions Chemistry Class 11th

B. O. D. value < 5 ppm for clean water and B.O.D value of polluted water 17 ppm.

Portland cement contains

Dicalcium silicate = 26%

Tricalcium silicate = 51%

Tricalcium aluminate = 11%

Hence major percentage is of tricalcium silicate

$\Delta H={\left(E_{act}\right)}_f-{\left(E_{act}\right)}_b$

= x – (x + y) = -y

              = -45 KJ/mol

              Ans. = 45

Applying : ${\left(n_1+n_2\right)}_{initial}={\left(n_1+n_2\right)}_{final}$  

Assuming the system attains a final temperature of T (Such that 300 < T < 60)

(Heat lost by N₂ of container l) = (Heat gained by N₂ of container II)

$n_{1}Cm\left(300-T\right)=n_{2}Cm\left(T-60\right)$  

$\left(\frac{2.8}{28}\right)\left(300-T\right)=\frac{0.2}{28}\left(T-60\right)$  

14 (300 – T) = T – 60

$P=\left(\frac{3}{28}\right)*8.31*\frac{284}{3*10^{-3}}*10^{-5}bar=0.84287bar$  

$P=84.28*10^{-2}bar$

$\approx 84*10^{-2}bar$  

Minimum temperature at which reaction becomes spontaneous is,

  $T_{min}=\frac{\Delta H^0}{\Delta S^0}$             

$\Delta {H^o}_{min}=\left[\Delta H_f^o\left(Fe\right)+\Delta H_f^o\left(CO\right)\right]-\left[\Delta H_f^o\left(FeO\right)+\Delta H_f^o\left(C\right)\right]$

$=\left[0-110.5\right]-\left[-266.3+0\right]$

$=155.8KJ/mol$

$\Delta S°=\left[\Delta S°\left(Fe\right)+\Delta S^o\left(CO\right)\right]-\left[\Delta S°\left(FeO\right)+\Delta S°\left(C\right)\right]$

$=\left(27.28+197.6\right)-\left(57.49+5.74\right)J/molK$

$=161.65J/molK$

$T_{min}=\frac{155.8*10^3}{161.65}K=963.8K\approx 964K$                        

Ionic radii of cations is smaller than anions, also more the positive charge less be the ionic radii and more the negative charge more be the ionic radii. Hence correct order of ionic radii is, $p^{3-}>S^{2-}>C{l}^{-}>K^{+}>C{a}^{2+}$

Millimoles of HCl = 200 * 0.2 = 40

Millimoles of NaOH = 300 * 0.1 = 30

Heat released =  $\frac{30}{1000}*57.1*1000J=1713J$

$\left[\begin{array}{l}\rho =\frac{m}{v}\\ m=\rho v\end{array}\right]$

Mass of solution = 500 * 1 g

= 500 g

Specific heat of water = 4.18 Jg^-1 K^-1

$\Delta T=\frac{q}{mc}$

$=\frac{1713}{500*4.18}°C$  

$=81.96*10^{-2}°C\approx 82*10^{-2}°C$

Ans. = 82

(1) Standard enthalpy of formation for alkali metal bromides becomes more negative on descending down the group.

(2) Standard enthalpy of formation for LiF is most negative among alkali metal fluorides.

(3) In case of Csl, lattice energy is less but Cs⁺ having less hydration energy due to which it is less soluble in water.

(4) For alkali metal fluorides, the solubility in water increases from Li to Cs. LiF is least soluble in water.

The Bond dissociation energy of D₂ is greater than H₂, therefore D₂ reacts slower than H₂.

$Unitof\frac{a{n}^2}{V^2}=atm$

$Unitofa=\frac{atm*unitof{V}^2}{unitof{n}^2}$

$=\frac{atm*{\left(d{m}^3\right)}^2}{mo{l}^2}$

= atm dm⁶ mol^-2